Chapter 18, Problem 5SP

(a)

Interpretation Introduction

Interpretation:

Balanced equation for the alpha decay of plutonium-239 has to be written.

Explanation of Solution

Alpha emission results in the formation of nucleus that has mass number less by 4 and atomic number less by 2.  Alpha particle is represented as $\text{H}\text{e}$.  Plutonium-239 has mass number of 239 and atomic number of 84.  After alpha emission, the particle formed will have mass number of 235 and atomic number of 82.  Element with atomic number of 82 is uranium.  Therefore, the balanced nuclear equation can be given as,

    $\text{P}\text{u}\stackrel{}{\to }\text{H}\text{e}\text{+}\text{U}$

(b)

Interpretation Introduction

Interpretation:

Mass of plutonium-239 present after 1000 year has to be calculated.

Explanation of Solution

Given mass of plutonium-239 is $\text{10}\text{.0}\text{g}$ and half-life of plutonium-239 is $\text{24,100}\text{years}$.

Decay rate constant can be calculated as shown below,

    $\begin{array}{l}k=\frac{0.693}{t_{1/2}}\\ =\frac{0.693}{24,100\text{yr}}\\ =2.8755\times {10}^{-5}{\text{yr}}^{-1}\end{array}$

The mass of plutonium-239 that remains after 1000 years can be calculated as shown below,

    $\begin{array}{l}\ln \left(\frac{\text{m}}{{\text{m}}_{\text{0}}}\right)=-kt\\ \ln \left(\frac{\text{m}}{\text{10}\text{.0}\text{g}}\right)=-(2.8755\times {10}^{-5}{\text{yr}}^{-1})(1000\text{yr)}\\ \text{=}-2.8755\times {10}^{-2}\\ \text{m}=10.0\text{g}\times e^{-0.028755}\\ =10.0\text{g}\times 0.9716\\ =9.716\text{g}\end{array}$

Therefore, mass of plutonium present after 1000 years is $\text{9}\text{.716}\text{g}$.

(c)

Interpretation Introduction

Interpretation:

Amount of energy given off by alpha emission of plutonium-239 has to be calculated.

Explanation of Solution

Change in mass for the alpha emission of plutonium-239 can be calculated as shown below,

    $\begin{array}{l}\Delta \text{m}={\text{m}}_{\text{reactants}}-{\text{m}}_{\text{products}}\\ ={\text{m}}_{{}^{\text{239}}\text{Pu}}-{\text{(m}}_{{}^{\text{235}}\text{U}}\text{+}{\text{m}}_{{}^{\text{4}}\text{He}})\\ \text{=}(\text{239}\text{.0521634 g/mol)}\\ -[\text{(235}\text{.0439299}\text{g/mol}\text{+}\text{4}\text{.001506}\text{g/mol)]}\\ \text{=}\text{0}\text{.0067275}\text{g/mol}\text{P}\text{u}\end{array}$

Nuclear binding energy is calculated as shown below,

$\begin{array}{l}{\text{E}}_{\text{b}}\text{=}{\text{(Δm)c}}^{\text{2}}\\ =\left(\text{0}\text{.0067275}\text{g/mol}\text{×}\frac{\text{1}\text{kg}}{\text{1000}\text{g}}\right)\times {\text{(2}\text{.99792}\text{×}{\text{10}}^{\text{8}}\text{m/s)}}^{\text{2}}\times \frac{1\text{J}}{1{\text{kgm}}^2{\text{s}}^{-2}}\times \frac{1\text{kJ}}{1000\text{J}}\\ =\frac{0.0604635}{1000000}\times {10}^{16}\text{kJ/mol}\\ \text{=}\text{6}\text{.04635}\times {10}^8\text{kJ/mol}\end{array}$

(d)

Interpretation Introduction

Interpretation:

Time required for the activity of plutonium-239 to decrease to $10.0\%$ of its initial activity has to be calculated.

Explanation of Solution

Given mass of plutonium-239 is $\text{10}\text{.0}\text{g}$ and half-life of plutonium-239 is $\text{24,100}\text{years}$.

Decay rate constant can be calculated as shown below,

    $\begin{array}{l}k=\frac{0.693}{t_{1/2}}\\ =\frac{0.693}{24,100\text{yr}}\\ =2.8755\times {10}^{-5}{\text{yr}}^{-1}\end{array}$

Initial activity is $100\%$ and final activity is $10.0\%$.

    $\begin{array}{l}\ln \left(\frac{\text{A}}{{\text{A}}_{\text{0}}}\right)=-kt\\ \ln \left(\frac{\text{10}\text{.0}\text{\%}}{\text{100}\text{\%}}\right)=-(2.8755\times {10}^{-5}{\text{yr}}^{-1})t\\ \ln (0.1)\text{=}(-2.8755\times {10}^{-5}y{r}^{-1})t\\ t=\frac{-2.3025}{-2.8755\times {10}^{-5}y{r}^{-1}}\\ =0.8007\times {10}^5\text{yr}\\ \text{=}\text{8}\text{.00}\times {\text{10}}^{\text{3}}\text{yr}\end{array}$

Time required is calculated as $\text{8}\text{.00}\times {\text{10}}^{\text{3}}\text{yr}$.

(e)

Interpretation Introduction

Interpretation:

Radiation that is received by $\text{70}\text{.0}\text{kg}$ person from a sample of plutonium-239 has to be calculated.

Explanation of Solution

Given mass of plutonium-239 is $\text{10}\text{.0}\text{g}$ and half-life of plutonium-239 is $\text{24,100}\text{years}$.

Decay rate constant can be calculated as shown below,

    $\begin{array}{l}k=\frac{0.693}{t_{1/2}}\\ =\frac{0.693}{24,100\text{yr}}\\ =2.8755\times {10}^{-5}{\text{yr}}^{-1}\end{array}$

First-order rate law can be used to determine the mass of plutonium-239 after $\text{30}\text{days}$ can be calculated as shown below,

    $\begin{array}{l}\ln \left(\frac{\text{m}}{{\text{m}}_{\text{0}}}\right)=-kt\\ \ln \left(\frac{\text{10}\text{.0}\text{g}\text{-}{\text{m}}_{\text{reacted}}}{\text{10}\text{.0}\text{g}}\right)=-(2.8755\times {10}^{-5}{\text{yr}}^{-1})(30\text{days)}\text{×}\frac{\text{1}\text{yr}}{\text{365}\text{.25}\text{days}}\\ =-2.36\times {10}^{-6}\end{array}$

    $\begin{array}{l}{\text{m}}_{\text{reacted}}\text{=}\text{10}\text{.0}\text{g}\text{-}\text{10}\text{.0}\text{g}\times {\text{e}}^{\text{-0}\text{.00000236}}\\ =2.36\times {10}^{-5}\text{g}\end{array}$

Energy that is released can be calculated as shown below,

    $\begin{array}{l}2.36\times {10}^{-5}\text{g}\times \frac{\text{1}\text{mol}}{\text{239}\text{.0521634}\text{g}}\text{×}\frac{\text{6}\text{.046}\text{×}{\text{10}}^{\text{8}}\text{kJ}}{\text{1}\text{mol}}=\frac{14.26856}{239.0521634}\times {10}^3\text{J}\\ \text{=}59.7\text{kJ}\end{array}$

The emission in rads can be calculated as shown below,

    $\begin{array}{l}\text{1}\text{rad}\text{=}\text{1}\text{.001}\text{×}{\text{10}}^{\text{-2}}\text{J/kg}\text{tissue}\\ \frac{\text{59}\text{.7}\text{kJ}}{\text{70}\text{.0}\text{kg}}\text{×}\frac{\text{1000}\text{J}}{\text{1}\text{kJ}}\text{×}\frac{\text{1}\text{rad}}{\text{1}\text{.001}\text{×}{\text{10}}^{\text{-2}}\text{J/kg}}=\frac{597}{70.07}\text{rad}\\ \text{=}\text{8}\text{.52}\times {10}^4\text{rad}\end{array}$

Effective dose in rem can be calculated using the quality factor as shown below,

    $\begin{array}{l}\text{Effective}\text{dose}\text{in}\text{rems}\text{=}\text{quality factor }\times \text{dose in rads}\\ =20\times \text{8}\text{.52}\times {10}^4\\ =1.704\times {10}^6\text{rems}\end{array}$

(f)

Interpretation Introduction

Interpretation:

The factors that has to be considered for burial and storage has to be commented.

Explanation of Solution

Time that is required to reduce the potency of plutonium-239 is thousands of years in macroscopic quantities.  Thus the burial and storage of plutonium-239 has to be done in a remote location that will not be used for any other purpose.  The burial unit has to be properly marked as dangerous and fenced.  The site has to be inaccessible to the general public.  This site also should have a minimal number of burrowing animal life and rooted plants.  Apart from this air testing and periodic soil testing has to be feasible.